According to Poiseuille’s law, the flow of liquid varies depending on the length of the tube, the radius of the tube, the pressure gradient and the viscosity of the fluid. It is a physical law that calculates the pressure drop in an incompressible Newtonian fluid flowing in laminar flow through a long cylindrical pipe with a constant cross-section. It further elaborates that the flow of a liquid is directly proportional to the pressure gradient and radius of the tube and inversely proportional to the fluid viscosity and length of the tube.

Formula

The flow of a liquid inside a pipe is denoted by the symbol Q. Its unit of measurement is the same as that of velocity, that is, m³/s. Its dimensional formula is given by [M⁰L³T^-1]. It is equal to the ratio of a pressure gradient to the resistance shown by the liquid flowing in a tube.

Q = ΔPπr⁴ / 8ηl

where,

ΔP is the pressure gradient,

π is a constant with the value of 3.14,

r is the radius of tube,

η is the viscosity of fluid,

l is the length of tube.

The formula can also be written as,

Q = ΔP/R

where,

ΔP is the pressure gradient,

R = 8ηl/πr⁴ is the resistance of the liquid.

Sample Problems

Problem 1. A liquid is flowing through a tube of radius 2 m and length of 3 m. If the pressure across the tube ends is 200 Pa, find the flow of liquid. The viscosity of the liquid is 0.051 Pa s.

Solution:

We have,

ΔP = 200

r = 2

η = 0.051

l = 3

Using the formula we get,

Q = ΔPπr⁴ / 8ηl

= (200 × 3.14 × 2⁴)/(8 × 0.051 × 3)

= (200 × 3.14 × 16)/(8 × 0.051 × 3)

= 10048/1.224

= 8213 m³/s

Problem 2. A liquid is flowing through a tube of radius 10 mm and length of 20 mm. If the pressure across the tube ends is 400 Pa, find the flow of liquid. The viscosity of the liquid is 0.0076 Pa s.

Solution:

We have,

ΔP = 400

r = 0.01

η = 0.0076

l = 0.02

Using the formula we get,

Q = ΔPπr⁴ / 8ηl

= (400 × 3.14 × 0.01⁴)/(8 × 0.0076 × 0.02)

= (400 × 3.14 × 10^-8)/(8 × 0.0076 × 0.02)

= 0.01035 m³/s

Problem 3. A liquid is flowing through a tube of radius 1 m and length of 0.5 m. If the flow of liquid is 5 m³/s, find the pressure across tube ends. The viscosity of liquid is 0.432 Pa s.

Solution:

We have,

Q = 5

r = 1

η = 0.432

l = 0.5

Using the formula we get,

Q = ΔPπr⁴ / 8ηl

=> 5 = (ΔP × 3.14 × 1⁴)/(8 × 0.432 × 0.5)

=> 3.14 ΔP = 8.64

=> ΔP = 2.75 Pa 

Problem 4. A liquid is flowing through a tube of radius 3 m and length of 8 m. If the flow of liquid is 10 m³/s, find the pressure across tube ends. The viscosity of the liquid is 0.0056 Pa s.

Solution:

We have,

Q = 10

r = 3

η = 0.056

l = 8

Using the formula we get,

Q = ΔPπr⁴ / 8ηl

=> 10 = (ΔP × 3.14 × 3⁴)/(8 × 0.056 × 8)

=> 254.34 ΔP = 35.84

=> ΔP = 0.14 Pa

Problem 5. A liquid is flowing through a tube with a pressure gradient of 100 Pa and a resistance of 200 Pa s/m³. Find the volumetric rate of flow.

Solution:

We have,

ΔP = 100

R = 200

Using the formula we get,

Q = ΔP/R

= 100/200

= 0.5 m³/s

Problem 6. A liquid is flowing through a tube with a pressure gradient of 75 Pa and resistance of 540 Pa s/m³. Find the volumetric rate of flow.

Solution:

We have,

ΔP = 75

R = 540

Using the formula we get,

Q = ΔP/R

= 75/540

= 0.138 m³/s

Problem 7. A liquid is flowing through a tube with a pressure gradient of 360 Pa. Find its resistance if its volumetric rate of flow is 2 m³/s.

Solution:

We have,

ΔP = 360

Q = 2

Using the formula we get,

Q = ΔP/R

=> R = ΔP/Q

=> R = 360/2

=> R = 180 Pa s/m³